Question

A right pyramid has regular octagon $ABCDEFGH$ with side length $1$ as its base and apex $V.$ Segments $\overline{AV}$ and $\overline{DV}$ are perpendicular. What is the square of the height of the pyramid?

$1$

$\dfrac{1 + \sqrt2}{2}$

$\sqrt2$

$\dfrac{3}{2}$

$\dfrac{2 + \sqrt2}{3}$

Solution:

Let $R$ be the circumradius of the octagon and $L$ the length of each lateral edge, so $L^2 = h^2 + R^2.$ Since $\angle AVD = 90^\circ,$ $AD^2 = 2L^2.$

Vertices $A$ and $D$ are three steps apart, a central angle of $135^\circ,$ so $AD^2 = 2R^2(1 - \cos 135^\circ) = R^2(2 + \sqrt2).$ Setting $R^2(2 + \sqrt2) = 2(h^2 + R^2)$ gives $2h^2 = R^2\sqrt2.$

For a regular octagon of side $1,$ $R^2 = \dfrac{1}{2\sin^2(22.5^\circ)} = \dfrac{2 + \sqrt2}{2}.$ Therefore $h^2 = \dfrac{R^2\sqrt2}{2} = \dfrac{(2 + \sqrt2)\sqrt2}{4} = \dfrac{2\sqrt2 + 2}{4} = \dfrac{1 + \sqrt2}{2}.$

Thus, the correct answer is B.

2024 AMC 12B Problem 23

Solution:

Problem 23 in Other Years